Question

Prove that the product of n geometric means between two quantities is equal to the nth power of a geometric mean of those two quantities.

Answer 1

\[\text { Let } G_1 , G_2 , G_3 , G_4 , . . . , G_n \text { be n G . M . s between a and b . }\]

\[\text { Then }, a, G_1 , G_2 , G_3 , G_4 , . . . , G_n , \text { b is a G . P .} \]

\[\text { Let r be the common ratio } . \]

\[ \because b = a_{n + 2} = a r^\left( n + 1 \right) \]

\[ \Rightarrow r = \left( \frac{b}{a} \right)^\frac{1}{\left( n + 1 \right)} \]

\[ \therefore G_1 = a_2 = ar\]

\[ G_2 = a_3 = a r^2 \]

\[ G_3 = a_4 = a r^3 \]

\[ G_n = a_\left( n + 1 \right) = a r^n \]

\[\text { Also, let G be the G . M . between a and b } . \]

\[ \therefore G^2 = ab\]

\[\text { Now }, G_1 \times G_2 \times G_3 \times G_4 \times . . . \times G_n = ar \times a r^2 \times a r^3 \times a r^4 \times . . . \times a r^n \]

\[ = a^n \times r^\left( 1 + 2 + 3 + 4 + . . . . . . + n \right) \]

\[ = a^n \times r^\left( \frac{n\left( n + 1 \right)}{2} \right) \]

\[ = a^n \times \left[ \left( \frac{b}{a} \right)^\frac{1}{\left( n + 1 \right)} \right]^\left( \frac{n\left( n + 1 \right)}{2} \right) \]

\[ = a^n \times \left( \frac{b}{a} \right)^\frac{n}{2} \]

\[ = a^\frac{n}{2} \times b^\frac{n}{2} \]

\[ = \left( ab \right)^\frac{n}{2} \]

\[ = \left( \sqrt{ab} \right)^n \]

\[ = G^n \]

\[ \therefore G_1 \times G_2 \times G_3 \times G_4 \times . . . \times G_n = G^n\]